Electrochemistry

Consider the following cell reaction

\(2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{I}^{-}(a q) \rightleftharpoons 2 \mathrm{Fe}^{2+}(a q)+\mathrm{I}_2(s)\)

At 298 K , the cell emf is 0.237 V . The equilibrium constant for the reaction is $10^x$. The value of $x$ is $\left(F=96500 \mathrm{C} \mathrm{mol}^{-1} ; R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$.